Question

(a+b)^2+(aw+bw^2)^2+(aw^2+bw)^2 =6ab write correct answer ​

Answer 1

Given Question

Prove that

$\sf \: {(a + b)}^{2} + {(a\omega + b {\omega }^{2})}^{2} + {(a {\omega }^{2} + b\omega )}^{2} = 6ab$

$\green{\large\underline{\sf{Solution-}}}$

Consider, LHS

$\rm :\longmapsto\:\sf \: {(a + b)}^{2} + {(a\omega + b {\omega }^{2})}^{2} + {(a {\omega }^{2} + b\omega )}^{2}$

$\sf \: = \: {a}^{2} + {b}^{2} + 2ab + {a}^{2} {\omega }^{2} + {b}^{2} {\omega }^{4} + 2(a\omega )(b {\omega }^{2}) + {a}^{2} {\omega }^{4} + {b}^{2} {\omega }^{2} + 2(b\omega )(a {\omega }^{2})$

$\sf \: = \: {a}^{2} + {b}^{2} + 2ab + {a}^{2} {\omega }^{2} + {b}^{2} {\omega }^{3}(\omega ) + 2ab{\omega }^{3}+ {a}^{2} {\omega }^{3}(\omega ) + {b}^{2} {\omega }^{2} + 2ab{\omega }^{3}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {\omega }^{3} = 0}}}$

So, using this, we get

$\sf \: = \: {a}^{2} + {b}^{2} + 2ab + {a}^{2} {\omega }^{2} + {b}^{2} \omega + 2ab+ {a}^{2}\omega+ {b}^{2} {\omega }^{2} + 2ab$

$\sf \: = \: {a}^{2}(1 + \omega + {\omega }^{2}) + {b}^{2}(1 + \omega + {\omega }^{2}) + 6ab$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ 1 + \omega + {\omega }^{2} = 0}}}$

So, using this, we get

$\sf \: = \: {a}^{2}(0) + {b}^{2}(0) + 6ab$

$\rm \:  =  \: 6ab$

Hence,

$\boxed{\tt{ \sf \: {(a + b)}^{2} + {(a\omega + b {\omega }^{2})}^{2} + {(a {\omega }^{2} + b\omega )}^{2} = 6ab}}$

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LEARN MORE : PROOF OF CUBE ROOTS OF UNITY

$\rm :\longmapsto\:x = {\bigg(1\bigg) }^{\dfrac{1}{3} }$

So, on cubing both sides, we get

$\rm :\longmapsto\: {x}^{3} = 1$

$\rm :\longmapsto\: {x}^{3} - 1 = 0$

$\rm :\longmapsto\:(x - 1)( {x}^{2} + x + 1)= 0$

$\rm\implies \:x = 1$

OR

$\rm\implies \: {x}^{2} + x + 1 = 0$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4(1)(1) } }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ 1 - 4}}{2}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{-3}}{2}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{3} \: i}{2}$

$\rm\implies \:x = \dfrac{ - 1 \: + \: \sqrt{3} \: i}{2} \: \: or \: \: x = \dfrac{ - 1 \: - \: \sqrt{3} \: i}{2}$

So, Cube roots of unity are

$\rm\implies \:x =1 \: \: or \: \: \dfrac{ - 1 \: + \: \sqrt{3} \: i}{2} \: \: or \: \: x = \dfrac{ - 1 \: - \: \sqrt{3} \: i}{2}$

where,

$\purple{\rm :\longmapsto\:\dfrac{ - 1 \: + \: \sqrt{3} \: i}{2} = \omega }$

and

$\purple{\rm :\longmapsto\:\dfrac{ - 1 \: - \: \sqrt{3} \: i}{2} = {\omega }^{2} }$

Now, Consider

$\purple{\rm :\longmapsto\:1 + \omega + {\omega }^{2}}$

$\sf \: = \: 1 + \dfrac{ - 1 \: + \: \sqrt{3} \: i}{2} + \dfrac{ - 1 \: - \: \sqrt{3} \: i}{2}$

$\sf \: = \: \dfrac{2 - 1 \: + \: \sqrt{3} \: i \: - \: 1 \: - \: \sqrt{3} \: i}{2}$

$\sf \: = \: 0$

Hence,

$\purple{\rm :\longmapsto\:\boxed{\tt{ 1 + \omega + {\omega }^{2} = 0}}}$

Now, Consider

$\purple{\rm :\longmapsto\: {\omega }^{3}}$

$\sf \: = \: \omega \times {\omega }^{2}$

$\sf \: = \: \dfrac{ - 1 \: + \: \sqrt{3} \: i}{2} \times \dfrac{ - 1 \: - \: \sqrt{3} \: i}{2}$

$\sf \: = \: \dfrac{ {( - 1)}^{2} - {( \sqrt{3}i) }^{2} }{4}$

$\sf \: = \: \dfrac{1 - {3i}^{2} }{4}$

$\sf \: = \: \dfrac{1 - 3( - 1)}{4}$

$\sf \: = \: \dfrac{1 + 3}{4}$

$\sf \: = \: \dfrac{4}{4}$

$\sf \: = \: 1$

Hence,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {\omega }^{3} = 0}}}$